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2 years ago

a 9 am math class has eight fewer students than the 11 am class. find the number of students in rsch class if both is 84

1 answer:

4 0

I'm assuming you mean the total of both is 84. So 9 am has 38 students and 11 am has 46 students I basically just divided total of 84 by 2, added half the difference (4) to the 11 am and subtracted half the difference (4) from the 9am

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What is the following quotient? StartFraction RootIndex 3 StartRoot 60 EndRoot Over RootIndex 3 StartRoot 20 EndRoot EndFraction

FrozenT [24]

Answer:

$\sqrt[3]{3}$

Step-by-step explanation:

We are required to simplify the quotient: $\dfrac{\sqrt[3]{60} }{\sqrt[3]{20}}$

Since the numerator and denominator both have the same root index, we can therefore say:

$\dfrac{\sqrt[3]{60} }{\sqrt[3]{20}} =\sqrt[3]{\dfrac{60} {20}}$

$=\sqrt[3]{3}$

The simplified form of the given quotient is $\sqrt[3]{3}$ .

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2 years ago

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A. Region C B. Region D C. Region A D. Region B

Fittoniya [83]

Answer:

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Step-by-step explanation:

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3 years ago

The price of a European tour includes stopovers at five cities to be selected from 15 cities. In how many ways can a traveler pl

serious [3.7K]

Answer:

The correct answer is D. 360,360.

Step-by-step explanation:

Number of cities as stopovers available for selection for an European tour = 15

A traveler has to select a total of five cities from these the given 15 cities for his tour and order is determined by the travel company.

Therefore the number of ways the traveler plan such a tour is given by $\left[\begin{array}{ccc}15\\5\end{array}\right]$ = 3003.

Now the order of the cities chosen by the travel company is given by 3003 × 5! = 360,360 ways.

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3 years ago

Kiran paid $54.00 for 3 tickets to go to a corn maze. At the same rate, what is the cost of 5 tickets to the maze? *

Shtirlitz [24]

Answer:90

Step-by-step explanation:

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3 years ago

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Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases: a. Central area 5 .

Flauer [41]

Answer:

a) "=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b) "=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c) "=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d) "=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e) "=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f) "=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

We will use excel in order to find the critical values for this case

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases:

a. Central area =.95, df = 10

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b. Central area =.95, df = 20

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c. Central area =.99, df = 20

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d. Central area =.99, df = 50

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e. Upper-tail area =.01, df = 25

For this case we need on the right tail 0.01 of the area and on the left tail we will have 1-0.01 = 0.99 , that means $\alpha =0.01$

We can use the following excel code:

"=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f. Lower-tail area =.025, df = 5

For this case we need on the left tail 0.025 of the area and on the right tail we will have 1-0.025 = 0.975 , that means $\alpha =0.025$

We can use the following excel code:

"=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

8 0

3 years ago